hdoj--1171--Number Sequence（KMP）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16532    Accepted Submission(s): 7283

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

 

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

 

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

 

Sample Input

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

 

Sample Output

6-1

 

Source

HDU 2007-Spring Programming Contest

#include<stdio.h>#include<string.h>int m,n;int next[1000005];int a[1000005],b[1000005];void getnext(int p[]){int j=0,k=-1;next[0]=-1;while(j<m-1){if(k==-1||p[k]==p[j]){k++;j++;next[j]=k;}else k=next[k];}}int kmp(int a[],int b[]){int i=0,j=0;getnext(b);//得到b的next数组 while(i<n){if(j==-1||a[i]==b[j]){i++;j++;}elsej=next[j];if(j==m)return i-m+1;//如果j==m说明b数组已经遍历到底，输出当前a数组元素的下标 }return -1;}int main(){int t;scanf("%d",&t);while(t--){int i;memset(next,0,sizeof(next));scanf("%d%d",&n,&m);for(i=0;i<n;i++)scanf("%d",&a[i]);for(i=0;i<m;i++)scanf("%d",&b[i]);printf("%d\n",kmp(a,b));}return 0;}